Question

A balloon filled with $39.1$ mol helium has a volume of $876\mathrm{L}$ at ${0.0}^{\circ }\mathrm{C}$ and $1.00$ atm pressure. The temperature of the balloon is increased to ${38.0}^{\circ }\mathrm{C}$ as it expands to a volume of $998\mathrm{L}$, the pressure remaining constant. Calculate $q,w$, and $\mathrm{\Delta }E$ for the helium in the balloon. (The molar heat capacity for helium gas is $20.8\mathrm{J}/{}^{\circ }\mathrm{C}\cdot$ mol. $)$

Short answer

The work done by the helium gas during expansion (w) is -12361.5 J, the heat absorbed by the gas (q) is 30963.44 J, and the change in internal energy of the helium gas (ΔE) is 18601.94 J.

Step-by-step solution

Determine initial and final conditions of the balloon

We are given the following initial and final conditions: Initial conditions: - moles (n) = 39.1 mol - volume (V₁) = 876 L - temperature (T₁) = 0.0°C = 273.15 K (convert to Kelvin) - pressure (P) = 1.00 atm Final conditions: - volume (V₂) = 998 L - temperature (T₂) = 38.0°C = 311.15 K (convert to Kelvin)

Calculate work done by the gas during expansion

Since the pressure remains constant during the expansion, we can calculate the work done by the gas using the formula: $w=-P\mathrm{\Delta }V$ where - P is the constant pressure (1.00 atm) - ΔV is the change in volume (V₂ - V₁) Convert pressure from atm to J/L using the conversion factor: 1 atm = 101.325 J/L: P = 1.00 atm × 101.325 J/L = 101.325 J/L Now calculate the change in volume: ΔV = V₂ - V₁ = 998 L - 876 L = 122 L Now plug these values into the equation for work: $w=-101.325\frac{\text{J}}{\text{L}}\times 122\text{L}=-12361.5\text{J}$

Calculate heat absorbed by the gas

To calculate the heat absorbed by the helium gas, we will use the molar heat capacity (Cp) and the equation: $q=n{C}_p\mathrm{\Delta }T$ where - n is the number of moles of helium (39.1 mol) - Cp is the given molar heat capacity of helium (20.8 J/deg C·mol) - ΔT is the change in temperature (T₂ - T₁) First, calculate the change in temperature: ΔT = T₂ - T₁ = 311.15 K - 273.15 K = 38 K Now plug the values into the equation for heat: $q=39.1\text{mol}\times 20.8\frac{\text{J}}{{}^{\circ }\text{C}\cdot \text{mol}}\times 38\text{K}=30963.44\text{J}$

Calculate change in internal energy

Now that we have calculated the work done by the gas during expansion (w) and the heat absorbed by the gas (q), we can calculate the change in internal energy (ΔE) using the equation: ΔE = q + w Plug the calculated values of q and w into the equation: ΔE = 30963.44 J - 12361.5 J = 18601.94 J

Final answers

- The work done by the gas during expansion (w) = -12361.5 J - The heat absorbed by the gas (q) = 30963.44 J - The change in internal energy of the helium gas (ΔE) = 18601.94 J

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental equation that relates the pressure, volume, temperature, and the amount (in moles) of an ideal gas. It is typically written as $PV=nRT$, where $P$ represents the pressure, $V$ is the volume, $n$ is the number of moles of gas, $R$ is the ideal gas constant, and $T$ is the absolute temperature in Kelvin.

When examining problems like the expansion of helium in a balloon, the ideal gas law helps us understand how gases behave under different conditions. Since the problem keeps the pressure constant, it focuses on the direct relationship between temperature and volume, which aligns with Charles’s Law, a specific case of the ideal gas law. Changes in one influence the other in a predictable way, which is crucial when calculating work done by the gas during expansion or compression.

In terms of application, students should always remember to convert temperature from degrees Celsius to Kelvin, as absolute temperature is a requirement for the calculations. Also, when the number of moles and pressure are constant, as in our helium balloon example, the ideal gas law simplifies our understanding of how volume and temperature are related.

Heat Capacity

Heat capacity, a concept central to thermochemistry, measures the amount of heat energy required to raise the temperature of a substance by one degree Celsius. It can be expressed as specific heat capacity, which is the heat capacity per unit mass, or as molar heat capacity, which is the heat capacity per mole of substance. In our helium balloon problem, we use molar heat capacity, denoted by $C_p$, since it is given per mole of gas.

The equation to calculate the heat absorbed or released by a gas during a temperature change is $q=n{C}_p\mathrm{△}T$, where $n$ refers to the number of moles, $C_p$ is the molar heat capacity, and $\mathrm{△}T$ is the temperature change. Comprehension of this concept allows students to understand that the greater the heat capacity, the more energy is required to change the temperature, indicating that different substances respond to heat differently.

Contextualizing Heat Capacity

When explaining the significance of heat capacity in thermodynamics, it is beneficial to consider real-world applications, such as climate studies where the heat capacity of oceans affects global temperature changes. In our helium balloon scenario, heat capacity allows us to compute the exact amount of heat absorbed by the helium as its temperature increases during expansion.

Internal Energy

Internal energy, denoted as $E$ or sometimes $U$, is the total energy contained within a system, encompassing kinetic energy due to molecular motion and potential energy resulting from molecular interactions. In the context of our exercise, the internal energy of a gas can change when the gas undergoes a temperature change or when work is performed on or by the gas.

The first law of thermodynamics, which is a conservation of energy principle, states that the change in internal energy of a system ($\mathrm{△}E$) is equal to the heat added to the system ($q$) plus the work done on the system ($w$): $$\mathrm{△}E=q+w$$.

It is essential for students to grasp that work done by the system (as in our balloon example) is considered negative because the system is expending energy. Conversely, work done on the system is positive. Understanding these conventions helps avoid common errors in sign when calculating the change in internal energy. With our helium balloon, when it expands and does work on the surroundings, the internal energy of the gas decreases, unless balanced out by heat added to the system.